Find all polynomials $P(x)$ and $Q(x)$ with real coefficients such that $$P(Q(x))=P(x)^{2017}$$for all real numbers $x$.
Problem
Source: Iran 3rd round 2017 first Algebra exam
Tags: algebra, polynomial, Iran, IranMO
m.yekta
31.08.2017 16:04
Amin12 wrote: Find all polynomials $P(x)$ and $Q(x)$ with real coefficients such that $$P(Q(x))=P(x)^{2017}$$for all real numbers $x$. It is quite obvious that $deg (Q(x))=2017$. let $P(x)=(x-z_1)^{a_1}...(x-z_n)^{a_n}$for some distinct $z_1,...,z_n$. We have $(Q(x)-z_1)^{a_1}...(Q(x)-z_n)^{a_n}=(x-z_1)^{2017a_1}...(x-z_n)^{2017a_n}$. $x=z_i \implies Q(z_i)=z_j,1\le i \le n$. if $Q(z)=z_i$ then $ x=z \implies z=z_j$(*). Suppose $Q(z_i)=z_j,Q(z_t)=z_j$, therefore the set $\{Q(z_1),...,Q(z_n)\}$ has less then $n-1$ elements. this is contradiction since we know there exists some $1 \le j \le n$ such that $Q(z_j)=z_i, 1\le i \le n$ using $(*)$. So we have $Q(z_i)\not= Q(z_j) \iff i \not = j. (**)$ Write $Q(x)=H(x)+z_k$ from $(*)$ we get $H(x)=(x-z_1)^{b_1}...(x-z_n)^{b_n}$ so $Q(x)=(x-z_1)^{b_1}...(x-z_n)^{b_n}+z_k$, if $b_i,b_j \not=0$then we have $Q(z_i)=Q(z_j)$, this is not true because of $(**)$. So we can write $Q(x)=(x-z_i)^{2017}+z_k$.We can do this for all $1\le i \le n$: $Q(x)=(x-z_1)^{2017}+z_{i_1}=(x-z_2)^{2017}+z_{i_2}=...=(x-z_n)^{2017}+z_{i_n} $ $\implies Q^{'}(x)=2017(x-z_1)^{2016}=2017(x-z_2)^{2016}=...=2017(x-z_n)^{2016} \implies z_1=z_2=...=z_n $ $\implies P(x)=(x-z)^{n} ,P(x) \in \mathbb {R}[x] \implies z\in \mathbb {R}$ $\boxed {Solution: P(x)=(x-r)^{n},Q(x)=(x-r)^{2017}+r}$
FISHMJ25
14.05.2019 01:02
m.yekta wrote: Amin12 wrote: Find all polynomials $P(x)$ and $Q(x)$ with real coefficients such that $$P(Q(x))=P(x)^{2017}$$for all real numbers $x$. It is quite obvious that $deg (Q(x))=2017$. let $P(x)=(x-z_1)^{a_1}...(x-z_n)^{a_n}$for some distinct $z_1,...,z_n$. We have $(Q(x)-z_1)^{a_1}...(Q(x)-z_n)^{a_n}=(x-z_1)^{2017a_1}...(x-z_n)^{2017a_n}$. $x=z_i \implies Q(z_i)=z_j,1\le i \le n$. if $Q(z)=z_i$ then $ x=z \implies z=z_j$(*). Suppose $Q(z_i)=z_j,Q(z_t)=z_j$, therefore the set $\{Q(z_1),...,Q(z_n)\}$ has less then $n-1$ elements. this is contradiction since we know there exists some $1 \le j \le n$ such that $Q(z_j)=z_i, 1\le i \le n$ using $(*)$. So we have $Q(z_i)\not= Q(z_j) \iff i \not = j. (**)$ Write $Q(x)=H(x)+z_k$ from $(*)$ we get $H(x)=(x-z_1)^{b_1}...(x-z_n)^{b_n}$ so $Q(x)=(x-z_1)^{b_1}...(x-z_n)^{b_n}+z_k$, if $b_i,b_j \not=0$then we have $Q(z_i)=Q(z_j)$, this is not true because of $(**)$. So we can write $Q(x)=(x-z_i)^{2017}+z_k$.We can do this for all $1\le i \le n$: $Q(x)=(x-z_1)^{2017}+z_{i_1}=(x-z_2)^{2017}+z_{i_2}=...=(x-z_n)^{2017}+z_{i_n} $ $\implies Q^{'}(x)=2017(x-z_1)^{2016}=2017(x-z_2)^{2016}=...=2017(x-z_n)^{2016} \implies z_1=z_2=...=z_n $ $\implies P(x)=(x-z)^{n} ,P(x) \in \mathbb {R}[x] \implies z\in \mathbb {R}$ $\boxed {Solution: P(x)=(x-r)^{n},Q(x)=(x-r)^{2017}+r}$ Why $P(x)=(x-z_1)^{a_1}...(x-z_n)^{a_n}$ ? It should be $P(x)=z(x-z_1)^{a_1}...(x-z_n)^{a_n}$ ?And why $H(x)=(x-z_1)^{b_1}...(x-z_n)^{b_n}$ I dont get it.
FISHMJ25
14.05.2019 01:27
And also problem says that $$P(Q(x))=P(x)^{2017}$$holds for real $x$ but you use that it holds for complex as well. What you are doing is equivalent to: if $R(x)$ and $G(x)$ have equal values for all real $x$ then they have for complex to.
Astaulphe
15.03.2020 17:02
This solution is substantially the same as the one above but hopefully clearer.
If $ P $ is constant, then $ P^{2017} - P = 0 $ from which we find $ P = 0, \pm 1 $.
Suppose $ P $ isn't constant.
Because two distinct polynomials cannot coincide on more values than their degrees,
$$ \forall x, P(Q(x)) = P(x)^{2017} \implies P \circ Q = P^{2017} $$Thus
$$ \deg Q = \frac {\deg P \circ Q} {\deg P} = \frac {\deg P^{2017}} {\deg P} = 2017 $$Let $ P = c(X - r_1)^{a_1} \dots (X - r_n)^{a_n} $ and $ R = \{r_1, \dots, r_n\} $ with $ |R| = n $.
If $ r \in R $, then
$$ c(Q(r) - r_1)^{a_1} \dots (Q(r) - r_n)^{a_n} = P(r)^{2017} = 0 $$so one of the $ Q(r) - r_i $ is $ 0 $, meaning that $ Q(r) \in R $.
If $ Q(x) \in R $, then
$$ P(x)^{2017} = P(Q(x)) = 0 $$so $ x \in R $.
If $ Q $ is not surjective from $ R $ to $ R $, then one of the $ Q - r_i $ has no roots in $ R $ but all of its roots in $ R $ while being of degree $ 2017 $. Absurd.
Thus $ Q $ is a bijection (it is a surjection and $ n < \infty $) from $ R $ to $ R $ and we may define $ Q^{-1} $ the inverse operation.
This implies that $ Q - r_i $ has only one complex root: $ Q^{-1}(r_i) $. That is, $ Q - r_i = d_i(X - Q^{-1}(r_i))^{2017} $.
It means that
$$ Q = d_1(X - Q^{-1}(r_1))^{2017} + r_1 = \dots = d_n(X - Q^{-1}(r_n))^{2017} + r_n $$Comparing the front coefficient gives $ d_1 = \dots = d_n $.
Comparing the second coefficient gives $ Q^{-1}(r_1) = \dots = Q^{-1}(r_n) $, that is $ r_1 = \dots = r_n $ after applying $ Q $ to every side.
This way, we get $ n = 1, R = \{r\}, Q^{-1}(r) = r, Q = d(X - r)^{2017} + r $ and finally $ P = c(X - r)^a $.
Substituting back,
$$ c^{2017}(X - r)^{2017a} = P^{2017} = P \circ Q = cd^a(X - r)^{2017a} $$so $ c = \pm \sqrt[2016]{d^a} $.
Solutions are $ (P, Q) = (-1, Q), (0, Q), (1, Q), (\pm c^a(X - r)^a, c^{2016}(X - r)^{2017}) $ with $ c \in \mathbb R_+ $ and $ Q \in \mathbb R[X] $.
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